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On the Truncated Kernel Function Jean-Marie De Koninck
Département de Mathématiques et de Statistique
Université Laval
Québec G1V 0A6
Canada
mailto:jmdk@mat.ulaval.cajmdk@mat.ulaval.ca

Ismaïla Diouf
Département de Mathématiques et d'Informatique
FST - Université Cheikh Anta DIOP
BP 5005, Dakar-Fann
Senegal
mailto:isma.diouf@gmail.comisma.diouf@gmail.com

Nicolas Doyon
Département de Mathématiques et de Statistique
Université Laval Québec G1V 0A6
Canada
mailto:nicodoyon77@hotmail.comnicodoyon77@hotmail.com

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Abstract:

We study properties of the truncated kernel function $\gamma_2$ defined on integers $n\ge 2$ by $\gamma_2(n)=\gamma(n)/P(n)$, where $\gamma(n)=\prod_{p\vert n}p$ is the well-known kernel function and $P(n)$ is the largest prime factor of $n$. In particular, we show that the maximal order of $\gamma_2(n)$ for $n\le x$ is $(1+o(1))x/\log x$ as $x\to \infty$ and that $\sum_{n\le x} 1/\gamma_2(n)= (1+o(1)) \eta x/\log x$, where $\eta=\zeta(2)\zeta(3)/\zeta(6)$. We further show that, given any positive real number $u<1$, $\lim_{x\to \infty} \frac 1x \char93 \{n\le x: \gamma_2(n)<x^u\}= \lim_{x\to \infty} \frac 1x \char93 \{n\le x: n/P(n) < x^u\} = 1-\rho(1/(1-u))$, where $\rho$ is the Dickman function. We also show that $n/P(n)$ can very often be much larger than $\gamma_2(n)$, namely by proving that, given any $c\in [1,\xi)$, where $\xi$ is the unique solution to $\xi\log 2 = \log(1+\xi)+\xi \log(1+1/\xi)$, then

$$\char93 \{n\le x: \gamma_2(n) \ge n/(c\log n)\} = o\left( \char93 \{n\le x: n/P(n) \ge n/(c\log n)\} \right) \qquad (x\to \infty).$$